Finding parallel unit vector

**benny92000** · November 25th 2011, 11:21 AM

Problem: A unit vector parallel to vector V = (2, -3, 6) is ?

Answer is (-.29, .43, -.85)

Thanks. I thought to find a parallel vector you have to multiply through by the same number, so I chose the answer (-2, 3, -6)

**Plato** · November 25th 2011, 11:38 AM

Originally Posted by benny92000

Problem: A unit vector parallel to vector V = (2, -3, 6) is ? Answer is (-.29, .43, -.85)

Find the length, $\|V\|=\sqrt{2^2+(-3)^2+6^2}=7$ .

Now divide $\left\langle {\frac{2}{7},\frac{{ - 3}}{7},\frac{6}{7}} \right\rangle$ .

There is your parallel unit vector.

**benny92000** · November 25th 2011, 12:33 PM

Ok, and can the length be -7, as well? As in divide through by -7 to find the right answer? That is what my book says, so I just wanted to verify this.

**Plato** · November 25th 2011, 12:41 PM

Originally Posted by benny92000

Ok, and can the length be -7, as well? As in divide through by -7 to find the right answer? That is what my book says, so I just wanted to verify this.

Length is never negative.
However, dividing by $-7$ also gives a unit vector parallel to the given vector.

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